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w^2+9w-248=0
a = 1; b = 9; c = -248;
Δ = b2-4ac
Δ = 92-4·1·(-248)
Δ = 1073
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(9)-\sqrt{1073}}{2*1}=\frac{-9-\sqrt{1073}}{2} $$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(9)+\sqrt{1073}}{2*1}=\frac{-9+\sqrt{1073}}{2} $
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